August 28th, 2015, 03:53 PM  #11 
Senior Member Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 390 Thanks: 70 
V8 Archie, I don't understand how you arrived at your concluding implication in your first response. Please spell it out for me. 
August 28th, 2015, 04:00 PM  #12 
Senior Member Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 390 Thanks: 70 
V8 Archie, what you did in your second response is NOT what I did in my original post. I subtracted a sequence from an identical sequence, namely 1/2 + 1/4 + 1/8 ... from 1/2 + 1/4 + 1/8 ... 
August 28th, 2015, 04:10 PM  #13 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,622 Thanks: 2611 Math Focus: Mainly analysis and algebra 
No, you subtracted the first term of one series from the second term of another, the second from the third, the third from the fourth, etc.. Anyway, the first example I gave did subtract equal terms from each other and also returned the wrong answer. 
August 28th, 2015, 04:13 PM  #14 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,622 Thanks: 2611 Math Focus: Mainly analysis and algebra  
August 28th, 2015, 04:28 PM  #15  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,084 Thanks: 848 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
S = 1 + 2 + 3 + 4 + ... 2S = 2(1 + 2 + 3 + 4 +...) = 2 + 4 + 6 + 8 +... 2S  S = (2 + 4 + 6 + 8 + ...)  (1 + 2 + 3 + 4 +...) 2S  S = 1 + (2  2)  3 + (4  4)  5 + (6  6)  7 +... = 1  3  5  7  ... Or you could do it v8archie's way and get S = 1/2. Always make sure your sums converge, and even then be wary of changing the order of your series. These can be very tricky to work with. Dan  
August 29th, 2015, 12:32 PM  #16  
Senior Member Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 390 Thanks: 70  Quote:
Secondly, I subtracted n from 2n to get n as the left side of (3). I still see no problem. Finally I subtracted $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ from the $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ on the right side of (2). This seems to be the problem for some of you. As far as I can see, in all your counter examples with nonconverging series, none of you has subtracted an identical series as has been done above. For that reason, I do not see them as true counter examples.  
August 29th, 2015, 01:46 PM  #17  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,084 Thanks: 848 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
August 29th, 2015, 03:05 PM  #18  
Senior Member Joined: Feb 2010 Posts: 704 Thanks: 138 
$\displaystyle n = \dfrac{3}{2}+\dfrac{9}{4}+\dfrac{27}{8}+ \ldots$ Quote:
Put yourself in for a Fields Medal. You've shown that a bunch of positive numbers adds to $\displaystyle 3$.  
August 29th, 2015, 03:18 PM  #19  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,622 Thanks: 2611 Math Focus: Mainly analysis and algebra  Quote:
$$\begin{aligned} && S &= \phantom{1+} 1  1 + 11+\cdots \\ &\text{multiply by $1$} & S &= 1+ 1  1 + 11+\cdots \\ &\text{subtract} & 2S &= \phantom{} 1 \end{aligned}$$  
August 29th, 2015, 04:16 PM  #20 
Senior Member Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 390 Thanks: 70 
Thank you, mrtwhs. I am convinced that you have provided a true counterexample. Your sarcasm is also cute. Last edited by Timios; August 29th, 2015 at 04:34 PM. 

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